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Math Help - First Order Non-Linear Differential Equation

  1. #1
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    First Order Non-Linear Differential Equation

    Question: Solve the Initial value problem:

     (5t^4y^{-3}+7t^6y^7)dt + (-3t^5y^{-4} + 7t^7y^6)dy = 0, y(1) = 1 .

    Express your answer in the form  F(t,y) = 2 , where  F(t,y) has no constant term.

     F(t,y)= __________________  =  2

    Attempt at the question:

    First of all I checked the terms to see if they were exact and they are not. I took the integral with respect to  t for  M(t,y) which is the first part of the equation and it came out to be  t^5y^{-3} + t^7y^7 + k(y) . Afterwards I took the partial with respect to y of that integral we just got which was:  -3t^5y^{-4} + 7t^6y^7 + k'(y) . I now would equate both these equations together and would cancel terms. The result gives me  k'(y) = 0 . So, my equation  F(t,y) = t^5y^{-3} + t^7y^7 . But my question now is, how do I solve using the initial values since my equation has both terms  t and  y ?

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  2. #2
    Behold, the power of SARDINES!
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    This is an implicitly defined curve. You are told that the solution should look like

    F(y,t)=2 this must hold for all y and t

    You have F(y,t)=\frac{t^5}{y^3}+t^7y^7=2=F(1,1) so the implict solution is

    \frac{t^5}{y^3}+t^7y^7=2 we can check this by taking the derivative

    5t^4y^{-3}-(3t^5y^{-4})\frac{dy}{dt}+7t^6y^7+(7t^7y^6)\frac{dy}{dt}=0

    (-3t^5y^{-4}+7t^7y^6)dy+(5t^4y^{-3}+7t^6y^7)dt=0
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  3. #3
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    Ok, so my answer that i reached is correct. Essentially, what does no constant term mean? I just want to clarify that for my understanding!
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